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Near a spatially extended Hopf bifurcation, a critical oscillatory normal mode has a slowly varying complex amplitude. For a selected travelling-wave branch, a method of multiple scales expansion takes the formHere the distance from onset is of order ${\epsilon }^2$, $v_0$ is the critical eigenvector, and $v_g$ is the group velocity. The comoving coordinate removes the first spatial derivative of the envelope. Projection at the first resonant order gives $A_T=\rho A+D_0{A}_{XX}-G_0|A{|}^{2}A$. Spatial translation and the temporal phase of the Hopf oscillation give a constant-phase symmetry $A\mapsto e^{i\theta }A$, which permits the cubic term $|A{|}^{2}A$ but forbids a generic $A^2$ term. The coefficients are complex because the envelope has both growth and frequency detuning. For a supercritical branch with positive real diffusion, real rescalings and a uniform phase rotation give the complex Ginzburg–Landau equationA subcritical Hopf branch requires higher saturation terms instead. If spatial reflection makes opposite travelling waves critical together, two coupled amplitude equations are required initially; the scalar equation applies to a selected single-wave branch when its competing amplitude is stable.

Distant boundaries can still determine which travelling pattern is observed. They can select its phase and wavenumber, reflect a travelling disturbance, inject a competing wave, or let a growing wave packet leave the domain. In particular, amplification in a comoving frame can be convective wave-packet instability rather than absolute wave-packet instability at a fixed laboratory point. A finite-domain outcome then need not reflect intrinsic bulk modulation dynamics. Periodic boundary conditions are the appropriate idealization when studying bulk instabilities without end selection or reflected-wave forcing: they eliminate physical end layers and make the allowed envelope Fourier modes discrete. They do not assert that real distant boundaries are always negligible. Choose a periodic length compatible with the carrier, fix the phase winding, and compare perturbations within that same periodic problem. On an open physical domain the group velocity and actual boundary conditions must be restored.

The optimum carrier wavenumber corresponds to the spatially uniform envelope $A_f=e^{-icT}$, often called a flat state. Its phase is arbitrary. Write $A=e^{-icT}(1+r)e^{i\varphi }$ with real small amplitude and phase disturbances. To first order,A Fourier mode $e^{\lambda T+ikX}$ therefore has matrix and dispersion relationThe trace is negative. The amplitude mode at $k=0$ has eigenvalue $-2$, while the phase mode is neutral because constant phase shifts preserve the solution. Put $D=1+bc$. For every nonzero real $k$, the determinant is positive if $D\ge 0$, whereas $D<0$ makes it negative for $0<k^2<-2D/(1+b^2)$. ThusThis is the Benjamin-Feir stability condition. At $D=0$, nonzero modes still decay, but the leading long-wave decay is fourth order rather than diffusive. On a periodic envelope interval of length $L$, the smallest nonzero wavenumber is $2\pi /L$, so strict stability of all allowed nonzero modes requiresA sufficiently short periodic domain can therefore exclude the unstable band even when the infinite-domain Benjamin-Feir stability condition fails.

The amplitude is damped while the phase is slow. For a slowly varying phase, the leading slaved amplitude is $r=-\frac{1}{2}\{b{\varphi }_{XX}+({\varphi }_X{)}^2\}+\cdots$. Substitution in the phase equation gives the diffusion term $D{\varphi }_{XX}$ and the nonlinear frequency shift $(c-b)({\varphi }_X{)}^2$. The latter also follows directly from a constant phase gradient: an envelope with gradient $q$ has amplitude squared $1-q^2$ and frequency $c+(b-c)q^2$. Constant phase symmetry rules out undifferentiated $\varphi$; translation invariance gives constant coefficients. The leading comoving Ginzburg-Landau equation also has spatial parity, so odd linear derivatives are absent at this order. Its neutral phase eigenvalue expands asThis coefficient follows by inserting a power series for $\lambda$ in the quadratic dispersion relation. Close to $D=0$, it becomes $E=(1+b^2)/2>0$ at leading order. The long-wave phase reduction of the complex Ginzburg-Landau equation is consequentlyThe fourth derivative regularizes negative phase diffusion. Terms omitted here have higher order in the long-wave expansion. First-order advection has been removed by the group-velocity frame; higher odd derivatives in a more general travelling-wave problem are beyond this leading amplitude approximation.

Take $D=-d$ with $d>0$ small. At the instability boundary $c=-1/b$, so $g=-(1+b^2)/b$ is nonzero. Set $\xi =\sqrt{d/E}X$, $\tau =d^{2}T/E$, and $\varphi =(d/g)\Phi$. At leading order the phase dynamics become the potential Kuramoto-Sivashinsky equation,With $u=-2{\Phi }_{\xi }$ this isThe variable $u$ is a phase-gradient disturbance, not the original physical field. Periodicity of the phase fixes ${\int }_0^{P}ud\xi =0$ for the zero-winding branch; the Kuramoto-Sivashinsky equation conserves this mean. The scaled period is $P=L\sqrt{d/E}$.

At fixed period $P$, the flat phase has growth rates ${\lambda }_n=k_n^2-k_n^4$, $k_n=2\pi n/P$. Its first loss of stability occurs at $P=2\pi$. To find what replaces it, put $k=2\pi /P$ and consider $P$ just above $2\pi$. After choosing an origin by translation, expand a small steady modulation as $u=a\sin (k\xi )+b_2\sin (2k\xi )+\cdots$. Projection of $-u{u}_{\xi }$ onto the first two harmonics yieldsSince ${\lambda }_2=-12$ at onset, this harmonic is slaved to $b_2=k{a}^2/(2{\lambda }_2)+\cdots$. The fundamental amplitude obeysThe primary periodic bifurcation of the Kuramoto-Sivashinsky equation is therefore supercritical. For $0<P-2\pi \ll 1$,The radial eigenvalue of this nonzero branch is $-2{\lambda }_1+O({\lambda }_1^2)<0$; all higher harmonics remain damped and the modulation's spatial phase is neutral. Thus the small finite-amplitude modulation is stable in the sense of orbital stability, at fixed mean and fixed period, despite the flat state's instability. The two signs of $a$ represent translations of the same modulation. In the potential formulation the spatial mean of $\Phi$ advances at $⟨({\Phi }_{\xi }{)}^2⟩$: the original wave acquires a nonlinear frequency correction rather than requiring a time-independent absolute phase.

For larger $P$, further linear modes become active and the small-amplitude argument no longer determines stability. A periodic modulation $\bar{u}$ must then be tested with the operator $v_{\tau }=-v_{\xi \xi }-v_{\xi \xi \xi \xi }-{\partial }_{\xi }(\bar{u}v)$ on mean-zero periodic disturbances. Its translation mode is always neutral, and secondary oscillatory or more complicated modulation can occur. Moreover, stability against perturbations of the same period does not prove stability against disturbances of a larger period; those require a separate sideband or Floquet theory calculation. The controlled conclusion near the first threshold is the existence of a stable small modulation family, not universal stability of all periodic Kuramoto-Sivashinsky solutions.

A mode has dimensionless damping ratio $k/{\alpha }_n$, and characteristic rootsIf $k/{\alpha }_n\ll 1$, the normal mode oscillates at ${\omega }_n={\alpha }_n+O(k^2/{\alpha }_n)$ with exponentially decaying envelope $e^{-kt}$. As $k\to 0$, the solution approaches the undamped wave equation response.

If $k/{\alpha }_n\gg 1$, the roots are real and negative, withThe mode has a fast transient and a slow nonoscillatory relaxation. After the fast transient, neglecting $y_{tt}$ gives the diffusion equation $y_t\simeq c^2{y}_{xx}/(2k)$ for these long-wavelength modes. At $k={\alpha }_n$, the repeated root produces $t{e}^{-kt}$: this is critical damping.

Thus the critical parameter is $k/{\alpha }_n=1$ for each mode, and the fundamental threshold isWhen $k<\pi c/l$ every mode is oscillatory. For any fixed, finite $k$, sufficiently large $n$ still gives oscillatory modes, so large damping of the lowest modes does not mean that every mode of the full string is overdamped.

Write the perturbation Stokes streamfunction as $\psi =\varphi (r)e^{ikz-i\omega t}$, with $\omega =kc$ and nonzero $k$. Its radial and axial velocity amplitudes are $u=-ik\varphi /r$ and $w={\varphi }^{\prime }/r$. These automatically obey incompressibility. Linearizing the radial and axial Euler equations, with pressure amplitude $P$, givesThusDifferentiating the first expression, the terms involving $U^{\prime }{\varphi }^{\prime }$ cancel. Equating the two expressions for $P^{\prime }$ proves the axisymmetric inviscid pipe stability equation:At the axis, finite smooth axisymmetric velocity requires $\varphi =O(r^2)$, after choosing the irrelevant constant in the streamfunction. At the wall, impermeability gives $\varphi (a)=0$. There is no tangential no-slip condition in this inviscid problem. A decoupled azimuthal disturbance satisfies $ik(U-c)u_{\theta }=0$ and does not introduce a growing discrete mode when $U\ne c$.

For real $k$, suppose $c=c_r+i{c}_i$ with $c_i\ne 0$, so division by $U-c$ is legitimate. Multiply the ordinary differential equation by ${\varphi }^{\ast }/[r(U-c)]$, integrate, and apply integration by parts. The endpoint terms vanish by the preceding regularity and wall conditions, leavingTaking its imaginary part yieldsA growing temporal mode therefore requires a sign change of $(U^{\prime }/r{)}^{\prime }$, rather than a sign change of $U^{\prime \prime }$ alone. This is the cylindrical counterpart of Rayleigh's inflection-point theorem. If this derivative vanishes identically, the positive first integral rules out a nontrivial growing real-wavenumber mode as well. Isolated zeros without a sign change do not make the second integral vanish for a nonzero regular mode.

For the parabolic Poiseuille flow, $U^{\prime }/r=-2U_0/a^2$ is constant. Under the specified exclusion of critical levels, division givesSetting $\varphi =rf$, or equivalently choosing $n=1$ in $r^{-n}\varphi$, turns this intoIt is the Modified Bessel differential equation of order one. Hence $\varphi =r[A{I}_1(kr)+B{K}_1(kr)]$. Since $r{K}_1(kr)$ tends to a nonzero constant at the axis, it gives a singular radial velocity and must be excluded. The admissible mode shape is thereforeFor $k=0$, direct solution gives $\varphi =C{r}^2$ after axis regularity, and the wall condition again forces the trivial solution. For nonzero $k$, the nonzero roots of $I_1$ are imaginary. Writing $j_{1,l}>0$ for the zeros of the ordinary Bessel function $J_1$, the evanescent potential modes of inviscid Poiseuille flow areTheir perturbation vorticity vanishes: ${\partial }_{z}u-{\partial }_{r}w=k^2\varphi /r-({\varphi }^{\prime }/r{)}^{\prime }=0$. These modes grow exponentially in one axial direction and decay in the other. They can describe end-forced or spatially localized potential fields, but neither sign gives a bounded normal mode on a whole infinite pipe. Moreover, the reduced radial equation contains no $c$: it supplies no temporal eigenvalue condition. An arbitrary imaginary part assigned to $\omega =kc$ here is consequently not evidence for a physical temporal instability.

The real-wavenumber calculation does exclude growing discrete axisymmetric modes of this inviscid profile. It does not establish every form of stability: singular neutral disturbances with $c$ in the range of $U$, transient dynamics, non-axisymmetric disturbances, and finite-amplitude transition require separate treatment. The exclusion $U\ne c$ specifically removes the inviscid continuous spectrum.

With viscosity, linearize the Navier-Stokes equations instead and eliminate pressure to obtain a fourth-order cylindrical counterpart of the Orr-Sommerfeld equation. Impose both $\varphi (a)=0$ and ${\varphi }^{\prime }(a)=0$, with appropriate regularity at the axis. The resulting eigenvalue problem involves the Reynolds number and determines complex frequencies, rather than leaving $c$ arbitrary. Viscosity supplies diffusion, no-slip boundary layers and critical-layer regularization, so the inviscid sign argument alone no longer decides its spectrum. That observation does not imply that viscous pipe flow must have a growing linear eigenmode; finite-amplitude pipe transition is a different question.

Let $p_1(z)$ denote the Eulerian perturbation of a fluid variable that is the pressure amplitude, and keep $\rho$ constant. The linearized Euler equations and incompressible flow constraint areThe last equation gives $u=i{w}^{\prime }/k$. Substituting it into the first yieldsDifferentiating this formula and using the vertical momentum equation produces the Rayleigh equation for inviscid shear flow:For a growing normal mode, $c$ is nonreal, so $U-c$ never vanishes on the real $z$ axis. In each open region the base velocity is constant or linear and $U^{\prime \prime }=0$; therefore $w^{\prime \prime }-k^{2}w=0$. Decay away from the layer selects $e^{-kz}$ above and $e^{kz}$ below, while both exponentials are retained inside. Conveniently normalize these asThese are the spatially localized discrete modes relevant to instability, rather than all possible generalized neutral disturbances at a critical layer in a shear flow.

At each corner $U$ is continuous and $U^{\prime }$ jumps. The two sides have the same normal velocity and normal stress: there is no membrane, density jump or singular external force supplying a pressure jump. Thus $w$ and $p_1$ must be continuous. The pressure formula proves the required continuity of $(c-U)w^{\prime }+U^{\prime }w$. This is pressure matching at a piecewise-linear shear interface; it does not require $w^{\prime }$ to be continuous. Equivalently integrating the Rayleigh equation for inviscid shear flow through a corner gives $(U-c)[w^{\prime }]=[U^{\prime }]w$.

Write $\alpha =kd$, $y=c/U_0$, and $E=e^{2\alpha }$. Continuity of normal velocity givesAt $z=d$, the outside derivatives are $w_{+}^{\prime }=-kA$, $U_{+}^{\prime }=0$, while the inside derivatives are $w_{-}^{\prime }=-kB+kEC$, $U_{-}^{\prime }=U_0/d$. Hence pressure matching reduces toSimilarly at $z=-d$, the outside derivative is $kD$ and the inside derivative is $-kEB+kC$. Matching givesThe two equations can be written asA nonzero mode therefore requiresThis is the dispersion relation for the unbounded piecewise-linear shear layer.

With the phase convention $e^{ik(x-ct)}$, positive $\mathrm{Im}c$ gives exponential temporal growth. Since the displayed $c^2$ is real, a growing/decaying pair occurs precisely whenFor $0<\alpha \le 1/2$, $e^{-2\alpha }>1-2\alpha$ by the strict exponential tangent inequality, so every such mode is unstable. For $\alpha \ge 1/2$, define $f(\alpha )=2\alpha -1-e^{-2\alpha }$. Its derivative $2+2e^{-2\alpha }$ is positive, while $f(1/2)<0$ and $f(1)>0$. There is consequently exactly one root in that interval, andAt the cutoff the discrete roots merge at $c=0$; above it the roots are real and have no exponential growth. The physical wavelengths in the unstable band satisfy ${\lambda }_x=2\pi /k>2\pi d/{\alpha }_s\simeq 9.83d$.

This is a finite-thickness Kelvin-Helmholtz instability: long disturbances roll up the layer and draw kinetic energy from its shear. The off-diagonal $e^{-2kd}$ terms describe coupling of the two vorticity discontinuities. Short disturbances decay across the layer before strongly coupling its edges and are neutral. Indeed at $kd\to \infty$ the two phase speeds approach the velocities $\pm U_0$ of the separated edges. At $kd\to 0$, expansion of the numerator gives $c^2/U_0^2\to -1$, recovering the vortex sheet result $c=\pm i{U}_0$; the growth rate then behaves as $k|U_0|$.

For the basic axisymmetric flow, ${\mathbf{u}}_0=\Omega R{\mathbf{e}}_{\varphi }$. The steady radial Euler equation for fluid motion gives $-{\Omega }^{2}R=-p_{0,R}/\rho$; the other components give $p_{0,z}=p_{0,\varphi }=0$. Thus, in the absence of an additional body force,The additive constant is undetermined without an imposed reference pressure; the rigid boundary does not prescribe a uniform boundary pressure. Its shape is fixed by the container rather than by a free-surface pressure condition.

Let $v_R,v_{\varphi },v_z$ be the small velocity amplitudes and $W=p_1/\rho$ the pressure amplitude. Linearizing the cylindrical Euler equations for an inviscid fluid about uniform rotation, the material derivative is ${\partial }_t+\Omega {\partial }_{\varphi }$. The intrinsic frequency for the stated phase convention is thereforeThe radial and azimuthal curvature terms supply the Coriolis acceleration terms, givingFor $s\ne 0,\pm 2\Omega$, solving these equations givesThe incompressible flow constraint isOn substitution, the terms proportional to $m\Omega W_R$ cancel. The remaining equation iswhich provesThis is the pressure equation for inertial modes in a rigid rotating sphere. The exceptional frequencies used in the divisions must instead be treated in the original momentum system; they are not needed for the mode below.

A fixed rigid sphere is impermeable, so its normal velocity vanishes:Using the solved amplitudes and multiplying by $s(s^2-4{\Omega }^2)/i$ givesEquivalently divide by ${\bar{\sigma }}^2$ for $\bar{\sigma }\ne 0$. Smoothness means tangential slip is allowed: there is no no-slip condition on these inviscid normal modes. Nor is there a condition $W=0$, since the wall supplies the required varying normal pressure.

For an axisymmetric oscillation, $m=0$ and $s=\sigma$. Write $\beta =1-4{\Omega }^2/{\sigma }^2$. For $W=z(A{R}^2+B{z}^2+C)$ the radial and vertical operators areThe interior pressure equation therefore requiresThe boundary equation becomesSubstituting $R^2=r_0^2-z^2$, this polynomial must vanish for every boundary height. Its constant and $z^2$ coefficients consequently giveTogether with the interior relation the latter condition is $(4+\beta )A=0$. A nontrivial nonsingular oscillatory solution needs $A\ne 0$: if $A=0$ and $\beta \ne 0$, the other equations force $B=C=0$. Hence $\beta =-4$, givingThe cubic axisymmetric inertial mode of a rigid sphere is explicitlyIts cubic pressure is regular at the axis and centre, and its reconstructed velocity is divergence free and tangent to the sphere by the preceding identities. The frequency lies within the inertial wave band $|\sigma |<2|\Omega |$: Coriolis acceleration, rather than compression or a moving surface, supplies the restoring dynamics. The amplitude $A$ is arbitrary in this linear problem.

The hydrostatic buoyancy anomaly is $b^{\prime }=f_0{\psi }_z$, and its leading adiabatic evolution is $D_g{b}^{\prime }/Dt+N_0^{2}w=0$. The small vertical velocity advects the background stratification; vertical advection of the perturbation buoyancy is higher order. ConsequentlyThe impermeability condition at a stationary lower boundary $z=\alpha y$ is $w=\alpha v$. Using $v_g={\psi }_x$ gives the topographic buoyancy boundary condition for quasi-geostrophic flowFor application at the reference plane, the actual small-height condition is $|\alpha |L/D\ll 1$. Slopes scaled by $D/L$ are of Rossby-number order, so this topographic contribution is retained alongside the small ageostrophic vertical velocity. The dimensional slope should not be compared to the Rossby number without also specifying the aspect ratio.

The basic quasi-geostrophic streamfunction is $\overline{\psi }=-\Lambda zy$. It gives $U=\Lambda z$, basic buoyancy gradient ${\overline{b}}_y=-f_0\Lambda$, and spatially constant interior quasi-geostrophic potential vorticity. Linearizing its conservation law therefore givesThe printed equation omits the shear-advection term. This cannot be omitted for general disturbances: an interior anomaly proportional to $\cos [k(x-\Lambda zt)]$ is materially conserved but not independent of time at a fixed location. The intended zero-interior-PV Eady model modes do have $q^{\prime }=0$ and satisfy both expressions; the ensuing mode calculation is valid in that subspace.

At either boundary, linearizing the geostrophic material derivative includes $v^{\prime }{\overline{\psi }}_{zy}=-\Lambda {\psi }_x^{\prime }$. LetThe two boundary equations areThese are exactly the respective bottom and top slope corrections to the rigid-boundary buoyancy condition for quasi-geostrophic waves.

Assume periodicity horizontally and first consider $q^{\prime }=0$. Multiply its defining equation by ${\psi }_x^{\prime }$ and integrate over the layer. The two horizontal Laplacian terms vanish by integration by parts; the remaining volume term after one vertical integration is an $x$-derivative. ThereforeFor nonzero $A_0,A_D$, the boundary equations giveThe top advection term integrates to zero. Subtracting proves conservation of the boundary pseudomomentum of a sloping Eady layer:The original PDF uses squared vertical derivatives and the upper slope $\gamma$ in the first denominator; these details are corrupted in the converted TeX. For arbitrary nonzero interior PV the same calculation instead gives $d\mathcal{P}/dt=(f_{0}s{)}^{-1}\int {\psi }_x^{\prime }{q}^{\prime }dV$, which need not vanish. For example, at an instant take ${\psi }^{\prime }=\cos kx+(z/D{)}^2\sin kx$, independent of $y$; the zonal average of ${\psi }_x^{\prime }{q}^{\prime }$ is $-ks/D^2$. Thus the boundary-only conservation statement also needs the zero-interior-PV qualification.

For an exponentially growing smooth normal mode, this qualification is automatic: the interior equation is $ik(\Lambda z-c)\hat{q}=0$, and a phase speed with nonzero imaginary part cannot equal the real basic velocity. Therefore $\hat{q}=0$. If $A_0{A}_D<0$, the conserved boundary quadratic form is definite, so a nonzero exponentially growing boundary amplitude is impossible. Instability consequently requiresfor $\Lambda \ne 0$. When one $A$ vanishes, the corresponding boundary derivative is simply advected and must vanish for a growing mode. The other boundary then yields a real single-edge-wave speed; if both vanish, the zero-interior-PV Neumann problem has no nonzero growing mode. These degenerate cases are neutral and should not be handled by dividing by zero.

For equal slopes, put $\zeta =z/D$, $C=c/(\Lambda D)$, $M=N_0|k|D/|f_0|$, and $a=N_0^2\alpha /(f_0\Lambda )-1$. The normal-mode interior equation is ${\Phi }_{\zeta \zeta }-M^2\Phi =0$ for the zero-interior-PV modes. ThusThe coefficient equations areSetting their determinant to zero, without dividing by $C$ or $a$, givesso the Eady model with parallel sloping boundaries hasFor positive $k,f_0$, $M$ is the printed $\mu$, and $a$ is the printed $\tilde{\alpha }$. The expression is even in the sign of that vertical scale. For $a\ge 0$ the radicand is at least $1/4$, so both speeds are real and these modes have no exponential growth. For equal slopes the earlier necessary condition is merely $a^2>0$, which is satisfied in many stable cases and is not sufficient. At $a=0$ the speeds are $0$ and $\Lambda D$; the invariant with nonzero denominators cannot be used, but the determinant still correctly gives neutral modes. No claim excluding transient amplification is needed.

Write $\mathbf{c}=(U,V{)}^T$ and $\Gamma =\left(\begin{array}{cc}{\gamma }_{11}& {\gamma }_{12}\\ {\gamma }_{12}& {\gamma }_{22}\end{array}\right)$. A well-posed anisotropic diffusion problem requires $\Gamma$ to be symmetric positive definite: ${\gamma }_{11}>0$ and $\Delta ={\gamma }_{11}{\gamma }_{22}-{\gamma }_{12}^2>0$. For the normal mode in the source the dispersion relation isThe group velocity vanishes when $\mathbf{c}-2i\Gamma \mathbf{k}=0$. ConsequentlyThe requested anisotropic absolute-instability threshold for positive diffusion isFor positive definite diffusion this is also sufficient: the anisotropic Gaussian heat kernel has exponential factor $\exp [\mu t-(\mathbf{x}-\mathbf{c}t{)}^T{\Gamma }^{-1}(\mathbf{x}-\mathbf{c}t)/(4t)]$ and a $t^{-1}$ prefactor, so its fixed-position growth rate is precisely $\mu -f$.

Let the prescribed speed be $q$ and let $d_{\min }\le d_{\max }$ be the positive eigenvalues of $\Gamma$. The Rayleigh quotient for ${\Gamma }^{-1}$ givesAbsence of absolute hydrodynamic instability for every direction therefore requiresThis is an upper bound on $\mu$, not a minimum required value. The source's last sentence reverses this monotonicity: reducing the growth parameter always helps, so there is no finite lower threshold guaranteeing absence. If it had asked for instability in every direction, the corresponding condition would be $\mu >q^2/(4d_{\min })$. Without the positive-definite assumption, negative diffusion directions make the forward problem ill posed; a singular diffusion tensor requires a separate transport analysis and does not support the displayed inverse-matrix formula.

Assume $\gamma >0$, the positive diffusivity requirement for the forward heat kernel. A normal mode $e^{ikx-i\omega t}$ of the linear complex Ginzburg-Landau equation givesFor real $k$, the maximum temporal growth rate is $\mu$. The Briggs-Bers criterion starts the inverse time Laplace transform above the temporal growth spectrum and follows the two spatial roots as its frequency contour is lowered. A contributing spatial pinch point occurs when roots continued from opposite sides of the spatial contour coalesce and prevent its deformation. An arbitrary algebraic saddle is not sufficient without this branch selection.

Here $D=D_k=0$ givesTo check the pinch, start at $\omega =i\Omega$ with $\Omega$ large. The roots are $k=-iU/(2\gamma )\pm i\sqrt{U^2+4\gamma (\Omega -\mu )}/(2\gamma )$ and initially lie in opposite half-planes. They meet at $\Omega =\mu -U^2/(4\gamma )$, giving the causal saddle. Positive imaginary absolute frequency means fixed-position exponential growth. ThereforeFor $\mu \le 0$ there is no positive temporal growth. At $\mu =U^2/(4\gamma )$ the fixed-position exponential rate is zero; its algebraic prefactor, derived below, decays. When $U=0$ the convective interval is empty.

The source specifies only real coefficients. For $\gamma <0$, temporal growth is unbounded at high real wavenumber, so the usual causal forward-diffusion construction is ill posed and the above criterion does not apply. For $\gamma =0$ the equation reduces to transport with amplification; its point impulse remains a moving delta, rather than a spreading heat kernel. These cases cannot be included by division by $\gamma$.

Set $a=2\pi G{\rho }_0$. The axisymmetric Poisson equation for Newtonian gravity integrates to $(R{\Phi }_R{)}^{\prime }=4\pi G{\rho }_{0}R$. Regularity on the axis fixes the integration constant, so ${\Phi }_R=aR$. Radial force balance in the rotating fluid givesWith the stated central pressure, $p_c={\rho }_0{a}^2/(4k^2)$, this is a perfect square:The free surface lies at the first zero of pressure. For the source's positive rotation parameter, $R_0=\sqrt{2\pi G{\rho }_0}/k$; allowing either rotation direction requires $|k|$ in the denominator. The inward effective gravity is ${\Phi }_R-R{\Omega }^2=R(a-k^2{R}^2)$, so $\boxed{g_{\mathrm{eff}}(R_0)=0}$. In particular $p_R(R_0)=0$. This is the critically rotating self-gravitating cylinder.

For a normal mode proportional to $e^{i\omega t+im\varphi }$, the background material derivative becomes $i\sigma$, where $\sigma =\omega +m\Omega (R)$. Linearizing the centrifugal term gives $-2\Omega u_{\varphi }$ in the radial equation. In the azimuthal equation the background velocity gradient and cylindrical geometric term together give $(2\Omega +R{\Omega }^{\prime })u_R=3\Omega u_R$. Since the interior mass density is constant, write $W=p^{\prime }/{\rho }_0+{\Phi }^{\prime }$. The two momentum equations and incompressible flow condition are consequentlyFor $m\ne 0$, the last two equations express the other amplitudes in terms of $u=u_R$:Differentiate the second expression using $R{\Omega }^{\prime }=\Omega$ and $R{\sigma }^{\prime }=m\Omega$, then substitute into the radial momentum equation. Before any division by $\sigma$, the result isAway from corotation this gives the required ordinary differential equationAt corotation the original coupled equations, or the undivided equation, must be used. A regular axisymmetric radial velocity would vanish by the incompressible flow condition; the requested dipolar calculation has $m=1$.